24P3Q05

5a)

$\displaystyle \begin{aligned}\text{pressure}&=\frac{\text{force}}{\text{area}} &=\frac{\text{weight of liquid}}{\text{cross-sectional area}} &=\frac{\text{mass of liquid}\times g}{\text{cross-sectional area}} &=\frac{(\rho V)g}{A} &=\frac{\rho (Ah)g}{A} &=h\rho g\end{aligned}$

5bi)

$\displaystyle h=\frac{V}{A}=\frac{370}{\pi {{(3.0)}^{2}}}=13.1\text{ cm}=0.131\text{ m}$

5bii)

$\displaystyle h\rho g=(0.131)(880)(9.81)=1.13\text{ Pa}$

5biii)

$\displaystyle U=\rho Vg=(880)(130\times {{10}^{-6}})(9.81)=1.122\text{ N}$

$\displaystyle \begin{aligned}T+U&=W T&=W-U &=3.32-1.122 &=2.20\text{ N}\end{aligned}$

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